0.3 kg of-10°C ice and 2.0 kg of 20°C water are combined in a thermally insulati
ID: 1575375 • Letter: 0
Question
0.3 kg of-10°C ice and 2.0 kg of 20°C water are combined in a thermally insulating container so that there is no heat transfer between the water-ice mix and its environment. The water and ice attain thermal equilibrium at the same final temperature, Tr Some of your answers below should involve the variable Tt> 0. properties of water specific heat (solid) specific heat (liquid) specific heat (gas) heat of fusion (at 0°C) heat of vaporization (at 100°C) 2,300,000 J/kg 2,100 J/(kg K) 4,200 J/(kg K) 2,000 J/(kg K) 330,000 J/kg a. What is thel heat transfer necessary to transform the ice from solid at -10°C to liquid at the final temperature, T? This calculation should comprise of three distinct processes.Explanation / Answer
Given
mass of ice m1 = 0.3 kg at T1 = -10 0C
mass of water m2 = 2 kg , at T2 = 20 0C
if the ice completely melt , then the heat required is
here heat gain by ice = heat lost by water
so that the both attain the same temperature Tf
M1*C*(273.15-263.15)+m1*L +M1*C*(Tf-0) = m2*C*(T-Tf)
0.3*2100(273.15-263.15)+0.3*330000+0.3*4200(Tf - 273.15) = 2*4200(293.15-Tf)
Tf = 279.46 k
a) Q = M1*L+M*C*(Tf-273.15) = 0.3*330000 + 0.3*4200*(279.46-273.15) J = 106950.6 J
b) heat trnsfer necessary to bring the water from 20 0c to Tf is
Q = m2*C*dT
Q = 2*4200(13.69) J
Q = 114996 J
c) d) trnsfer of thermal energy into the object is to the ice and transfer thermal energy out of the object is from water
because water loosing energy and same time ice gaining thermal energy becomes water
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.